Rozkład pobierania próbek z dwóch niezależnych populacji Bernoulli

Załóżmy, że mamy próbki dwóch niezależnych zmiennych losowych Bernoulliego, $\mathrm{Ber}(\theta_1)$ i $\mathrm{Ber}(\theta_2)$ .

Jak udowodnimy, że

\frac{({\bar{X}}_{1} - {\bar{X}}_{2}) - (θ_{1} - θ_{2})}{\sqrt{\frac{θ_{1} (1 - θ_{1})}{n_{1}} + \frac{θ_{2} (1 - θ_{2})}{n_{2}}}} \overset{d}{\to} N (0, 1)

$\frac{(\bar X_1-\bar X_2)-(\theta_1-\theta_2)}{\sqrt{\frac{\theta_1(1-\theta_1)}{n_1}+\frac{\theta_2(1-\theta_2)}{n_2}}}\xrightarrow{d} \mathcal N(0,1)$ ?

Załóżmy, że $n_1\neq n_2$ .

distributions sampling bernoulli-distribution

— Stary człowiek na morzu.
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Z_i = X_1i - X_2i jest sekwencją iid rv o skończonej średniej i wariancji. Dlatego spełnia centralne twierdzenie graniczne Levy-Linderberga, z którego wynikają twoje wyniki. A może prosi o dowód samego clt?

— Three Diag

@ThreeDiag Jak stosujesz wersję LL CLT? Nie sądzę, że to prawda. Napisz odpowiedź, żeby sprawdzić szczegóły.

— Stary człowiek na morzu.

Wszystkie szczegóły już tam są. Aby zastosować LL, potrzebujesz sekwencji iid rv ze skończoną średnią i wariancją. Zmienna Z_i = X_i1 i X_i2 spełnia wszystkie trzy wymagania. Niezależność wynika z niezależności dwóch oryginalnych bernoulli i można zauważyć, że E (Z_i) i V (Z_i) są skończone poprzez zastosowanie standardowych właściwości E i V

— Three Diag

„próbki dwóch niezależnych zmiennych losowych Bernoulliego” - niepoprawne wyrażenie. Musi być: „dwie niezależne próbki z dystrybucji Bernoulliego”.

— Viktor,

Please add "as

n_{1}, n_{2} \to \infty

$n_1,n_2\to \infty$ ".

— Viktor

Odpowiedzi:

Put $a=\frac{\sqrt{\theta_1(1-\theta_1)}}{\sqrt{n_1}}$ , $b=\frac{\sqrt{\theta_2(1-\theta_2)}}{\sqrt{n_2}}$ , $A=(\bar{X}_1-\theta_1)/a$ , $B=(\bar{X}_2-\theta_2)/b$ . We have $A\to_d N(0,1),\ B\to_d N(0,1)$ . In terms of characteristic functions it means

ϕ_{A} (t) \equiv E e^{i t A} \to e^{- t^{2} / 2}, ϕ_{B} (t) \to e^{- t^{2} / 2} .

$\phi_A(t)\equiv {\bf E}e^{itA}\to e^{-t^2/2},\ \phi_B(t)\to e^{-t^2/2}.$ We want to prove that

D := \frac{a}{\sqrt{a^{2} + b^{2}}} A - \frac{b}{\sqrt{a^{2} + b^{2}}} B \to_{d} N (0, 1)

$D:=\frac{a}{\sqrt{a^2+b^2}}A-\frac{b}{\sqrt{a^2+b^2}}B\to_d N(0,1)$

Since $A$ and $B$ are independent,

ϕ_{D} (t) = ϕ_{A} (\frac{a}{\sqrt{a^{2} + b^{2}}} t) ϕ_{B} (- \frac{b}{\sqrt{a^{2} + b^{2}}} t) \to e^{- t^{2} / 2},

$\phi_D(t)=\phi_A\left(\frac{a}{\sqrt{a^2+b^2}}t\right)\phi_B\left(-\frac{b}{\sqrt{a^2+b^2}}t\right)\to e^{-t^2/2},$ as we wish it to be.

This proof is incomplete. Here we need some estimates for uniform convergence of characteristic functions. However in the case under consideration we can do explicit calculations. Put $p=\theta_1,\ m=n_1$ .

\begin{aligned} ϕ_{X_{1, 1}} (t) & = 1 + p (e^{i t} - 1), \\ ϕ_{{\bar{X}}_{1}} (t) & = (1 + p (e^{i t / m} - 1))^{m}, \\ ϕ_{{\bar{X}}_{1} - θ_{1}} (t) & = (1 + p (e^{i t / m} - 1))^{m} e^{- i p t}, \\ ϕ_{A} (t) & = (1 + p (e^{i t / \sqrt{m p (1 - p)}} - 1))^{m} e^{- i p t \sqrt{m} / \sqrt{p (1 - p)}} \\ = {((1 + p (e^{i t / \sqrt{m p (1 - p)}} - 1)) e^{- i p t / \sqrt{m p (1 - p)}})}^{m} \\ = {(1 - \frac{t^{2}}{2 m} + O (t^{3} m^{- 3 / 2}))}^{m} \end{aligned}

$\begin{align} \phi_{X_{1,1}}(t) &= 1+p(e^{it}-1), \\ \phi_{\bar X_{1}}(t) &= (1+p(e^{it/m}-1))^m, \\ \phi_{\bar X_{1}-\theta_1}(t) &= (1+p(e^{it/m}-1))^m e^{-ipt}, \\ \phi_{A}(t) &= (1+p(e^{it/\sqrt{mp(1-p)}}-1))^m e^{-ipt\sqrt{m}/\sqrt{p(1-p)}} \\[5pt] &= \left( \left(1+p(e^{it/\sqrt{mp(1-p)}}-1)\right)e^{-ipt/\sqrt{mp(1-p)}}\right)^m \\[5pt] &=\left( 1-\frac{t^2}{2m}+O(t^3m^{-3/2}) \right)^m \end{align}$ as

t^{3} m^{- 3 / 2} \to 0

$t^3m^{-3/2}\to 0$ . Thus, for a fixed

t

$t$ ,

ϕ_{D} (t) = {(1 - \frac{a^{2} t^{2}}{2 (a^{2} + b^{2}) n_{1}} + O (n_{1}^{- 3 / 2}))}^{n_{1}} {(1 - \frac{b^{2} t^{2}}{2 (a^{2} + b^{2}) n_{2}} + O (n_{2}^{- 3 / 2}))}^{n_{2}} \to e^{- t^{2} / 2}

$\phi_D(t)=\left( 1-\frac{a^2t^2}{2(a^2+b^2)n_1}+O(n_1^{-3/2}) \right)^{n_1} \left( 1-\frac{b^2t^2}{2(a^2+b^2)n_2}+O(n_2^{-3/2}) \right)^{n_2} \to e^{-t^2/2}$ (even if

a \to 0

$a\to 0$ or

b \to 0

$b\to 0$ ), since

| e^{- y} - (1 - y / m)^{m} | \leq y^{2} / 2 m

$\left|e^{-y}-(1-y/m)^m\right|\le {y^2}/{2m}\$ when

y / m < 1 / 2

$\ y/m<1/2$ (see /math/2566469/uniform-bounds-for-1-y-nn-exp-y/ ).

Note that similar calculations may be done for arbitrary (not necessarily Bernoulli) distributions with finite second moments, using the expansion of characteristic function in terms of the first two moments.

— Viktor
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This seems correct. I'll get back to you later on, when I have time to check everything. ;)

— An old man in the sea.

-1

Proving your statement is equivalent to proving the (Levy-Lindenberg) Central Limit Theorem which states

If $\{Z_i\}_{i=1}^n$ is a sequence of i.i.d random variable with finite mean $\mathbb{E}(Z_i) = \mu$ and finite variance $\mathbb{V}(Z_i) = \sigma^2$ then

\sqrt{n} (\bar{Z} - μ) \to^{d} N (0, σ^{2})

$\sqrt{n}(\bar{Z} - \mu) \to^d N(0,\sigma^2)$

Here $\bar{Z} = \sum_i Z_i/n$ that is the sample variance.

Then it is easy to see that if we put

Z_{i} = X_{1} i - X_{2} i

$Z_i = X_1i - X_2i$ with

X_{1 i}, X_{2 i}

$X_{1i}, X_{2i}$ following a

B e r (θ_{1})

$Ber(\theta_1)$ and

B e r (θ_{2})

$Ber(\theta_2)$ respectively the conditions for the theorem are satisfied, in particular

E (Z_{i}) = θ_{1} - θ_{2} = μ

$\mathbb{E}(Z_i) = \theta_1 - \theta_2 = \mu$

and

V (Z_{i}) = θ_{1} (1 - θ_{1}) + θ_{2} (1 - θ_{2}) = σ^{2}

$\mathbb{V}(Z_i)= \theta_1(1-\theta_1) +\theta_2(1-\theta_2)= \sigma^2$

(There's a last passage, and you have to adjust this a bit for the general case where $n_1 \neq n_2$ but I have to go now, will finish tomorrow or you can edit the question with the final passage as an exercise )

— Three Diag
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I could not obtain what I wanted exactly because of the possibility of

n_{1} \neq n_{2}

$n_1\neq n_2$

— An old man in the sea.

I will show later if you can't get it. Hint: compute the variance of the sample mean of Z and use that as the variable in the theorem

— Three Diag

Three, could you please add the details for when

n_{1} \neq n_{2}

$n_1 \neq n_2$ ? Thanks

— An old man in the sea.

Will do as soon as find a little timr. There was in fact a subtlety that prevents from using LL clt without adjustment. There are three ways to go, the simplest of which is invoking the fact that for large n1 and n2, X1 and X2 go in distribution to normals, then a linear combination of normal is also normal. This is a property of normals that you can take as given, otherwise you can prove it by characteristic functions.

— Three Diag

The other two require either a different clt (Lyapunov possibly) or alternatively treat n1 = i and n2= i +k. Then for large i you can essentially disregard k and you can go back to apply LL (but still it will require some care to nail the right variance)

— Three Diag